Monostable Multivibrator using an Op-amp:

As shown C₁ R₁ and D₂ constitute a pulse shaping circuit. A monostable circuit, as the name implies, contain a permanent stable state. It may be easily deduced that this permanent stable state is V₀ = V_sat = HIGH (with D₁ clamped at 0.6 V, if β is sufficient to make + β V_sat > 0.6, the circuit would be in V₀ = + V_sat = HIGH state. Even if the circuit accidentally begin from V₀ = - V_sat, signal at non-inverting input is - β V_sat and capacitor C₀ charges until voltage across it becomes more negative than - β V_sat at which time V₀ switches to + V_sat and would remain so for all times to come until forced to modify by applying an external trigger.

Upon the application of an external signal (V_e) resulting in a short (narrow) negative going trigger pulse (V_tr) at t = 0, the V₀ goes LOW. V_co changes exponentially towards - V_sat and when V_co attain more negative than - β V_sat, output becomes HIGH. The time- period during which waveform remains LOW (- V_sat) may be followed to be time taken by C₀ to change from 0.6 volts to β V_sat and may be calculated from

                    V_co (t ) = A + B e  ^{(- t / T1)}

 Considering that with t = 0, V_co (0) = 0.6 and for t = ∞ , V_co ( ∞ ) = - V_sat, one may find

that

                                 V_co (t ) =- V_sat  + (V_sat  + 0.6) e^{( - t /T1)}

For t = T, V_co = - β V_sat. Therefore, we determine

                         T = C₀  R_F    ln  (V _sat + 0.6 )/ (1 - β) V_sat )

Figure: Monostable Multivibrator using an Op-amp Comparator